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prisoha [69]
3 years ago
14

The following transition occurs at a molecular level for a substance. What transition corresponds to this change in microscopic

structure? The carbon dioxide molecules on the left are in a regular, tightly packed pattern. After heating, it becomes much lower density. A. Melting B. Boiling C. Sublimation D. Freezing
Chemistry
2 answers:
olchik [2.2K]3 years ago
5 0

Answer:

D. Freezing

Explanation:

Please mark brainliest and have a great day!

Gnoma [55]3 years ago
3 0

Answer:

A. Melting

Explanation:

In the above question we noticed that there is a passage from a regular and well compacted carbon dioxide molecules to molecules in a less compact and lower density pattern after a period of heating. According to this description we can conclude that a physical process called "Melting" occurred, which is the process, where a solid substance (which has very compacted molecules) when heated becomes a liquid substance, with less compacted molecules and lower density.

Melting explains the changeover from heating to solid state to liquid state. Heating causes the temperature of the substance to rise to its melting point. The temperature does not rise while fusion is taking place, that is, only after all the substance has gone into liquid state does the temperature rise again.

The melting point of a substance is the temperature at which that substance changes from solid to liquid state.

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The equilibrium constant in terms of pressure. Kp for the following process is 0.179 at 50 °C. It increases to 0.669 at 86 °C. C
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Answer:

a) ΔHvap=35.3395 kJ/mol

b) Tb=98.62 °C

Explanation:

Given the reaction:

C₇H₁₆ (l) ⇔ C₇H₁₆ (g)

Kp=P(C₇H₁₆) since the concentration ratio for a pure liquid is equal to 1.

When

T₁=50°C=323.15K ⇒P₁=0.179

T₂=86°C=359.15K ⇒P₂=0.669

The Clasius-Clapeyron equation is:

ln(\frac{P_2}{P_1}) =-\frac{AH_{vap}}{R} (\frac{1}{T_2}-\frac{1}{T_1})

ln(\frac{0.669}{0.179}) =-\frac{AH_{vap}}{8.3145 J.mol^{-1}K^{-1}} (\frac{1}{359.15K}-\frac{1}{323.15K})

1.3184 =-\frac{AH_{vap}}{8.3145 J.mol^{-1}K^{-1}} (-3.10186*10^{-4}K{^-1})

ΔHvap=35339.5 J/mol=35.3395 KJ/mol

Normal boiling point ⇒ P=1 atm

Hence, we find the normal boiling point where:

T₁=323.15K

P₁=0.179 atm

P₂=1 atm

ln(\frac{P_2}{P_1}) =-\frac{AH_{vap}}{R} (\frac{1}{T_2}-\frac{1}{T_1})

ln(\frac{1atm}{0.179atm}) =-\frac{35339.5 J/mol}{8.3145 J.mol^{-1}K^{-1}} (\frac{1}{T_2}-\frac{1}{323.15K})

1.7203=-4250.34 (\frac{1}{T_2}-\frac{1}{323.15K})

T₂=371.77 K= 98.62 °C

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